NUST NET Physics chapter 4: Physics · Heat and thermodynamics MCQs with answers

By Team Megalecture · Updated 4 Nov 2026

14 exam-style questions on Physics · Heat and thermodynamics, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.

14 questions · 1 topics · 10 per test

Test this chapter freeAll NUST NET Physics chapters

What chapter 4 covers

Ten Physics · Heat and thermodynamics questions with answers

Try each one, then open the answer. These ten stay the same; the test draws new ones every time.

1. An ideal gas occupies 2.0 L at 27 °C. It is heated at constant pressure to 177 °C. Its volume becomes

  1. A
    1.3 L
  2. B
    3.0 L
  3. C
    2.0 L
  4. D
    13 L
Show answer

Answer: B. At constant pressure V ∝ T in kelvin: V = 2.0 × 450/300 = 3.0 L. Using Celsius gives 2.0 × 177/27 ≈ 13 L.

2. A diatomic ideal gas (Cv = 5R/2) is heated at constant pressure. What fraction of the heat supplied goes into raising its internal energy?

  1. A
    2/7
  2. B
    2/5
  3. C
    1
  4. D
    5/7
Show answer

Answer: D. Q = nCpΔT with Cp = Cv + R = 7R/2, and ΔU = nCvΔT, so ΔU/Q = 5/7. 2/7 is the fraction that does work.

3. For a monatomic ideal gas Cv = 3R/2. The ratio γ = Cp/Cv is

  1. A
    3/2
  2. B
    7/5
  3. C
    4/3
  4. D
    5/3
Show answer

Answer: D. Cp = Cv + R = 5R/2, so γ = (5R/2) ÷ (3R/2) = 5/3. 7/5 is the value for a diatomic gas.

4. The temperature of a gas rises from 27 °C to 327 °C. The average translational kinetic energy of its molecules

  1. A
    stays the same
  2. B
    increases √2 times
  3. C
    doubles
  4. D
    increases about 12 times
Show answer

Answer: C. Average KE = (3/2)kT depends on kelvin temperature, which goes from 300 K to 600 K, so it doubles. √2 is the factor for the rms speed; 12 uses Celsius.

5. When a gas is heated at constant volume,

  1. A
    all the heat supplied increases its internal energy
  2. B
    all the heat supplied is used to do external work
  3. C
    its internal energy stays constant
  4. D
    its temperature stays constant
Show answer

Answer: A. With no change in volume the gas does no work (W = PΔV = 0), so the first law gives Q = ΔU: all the heat raises the internal energy and hence the temperature.

6. According to the kinetic theory, the pressure of an ideal gas equals

  1. A
    two-thirds of the average translational kinetic energy of its molecules per unit volume
  2. B
    the average kinetic energy of one molecule
  3. C
    one-third of the total kinetic energy of all its molecules
  4. D
    three-halves of its kinetic energy per unit volume
Show answer

Answer: A. P = ⅓ρ⟨v2⟩ = ⅔ (N/V)⟨½mv2⟩, two-thirds of the translational KE per unit volume. Total KE alone has the wrong unit for pressure.

7. A gas in a sealed rigid container is at 2.0 atm and 27 °C. It is heated to 327 °C. Its pressure becomes

  1. A
    1.0 atm
  2. B
    4.0 atm
  3. C
    24 atm
  4. D
    2.0 atm
Show answer

Answer: B. At constant volume P ∝ T in kelvin: P = 2.0 × 600/300 = 4.0 atm. Using Celsius temperatures gives 2.0 × 327/27 ≈ 24 atm.

8. When an ideal gas expands isothermally,

  1. A
    its internal energy stays constant and the heat it absorbs equals the work it does
  2. B
    it absorbs no heat from its surroundings
  3. C
    its internal energy falls by the amount of work it does
  4. D
    its temperature falls
Show answer

Answer: A. For an ideal gas U depends only on T, so ΔU = 0 and by the first law Q = W. Falling internal energy with no heat flow describes an adiabatic expansion.

9. A gas is compressed adiabatically, 200 J of work being done on it. Its internal energy

  1. A
    falls by 200 J
  2. B
    stays the same
  3. C
    rises by 200 J
  4. D
    rises by 400 J
Show answer

Answer: C. Adiabatic means Q = 0, so ΔU equals the work done on the gas: it rises by 200 J and the gas warms up.

10. The rms speed of the molecules of a gas at 27 °C is v. At what temperature will it be 2v?

  1. A
    54 °C
  2. B
    327 °C
  3. C
    1200 °C
  4. D
    927 °C
Show answer

Answer: D. vrms ∝ √T, so doubling the speed needs 4 × 300 K = 1200 K = 927 °C. 1200 °C forgets to convert back from kelvin, and 327 °C (600 K) only doubles T.

Start the chapter testAbout the NUST NET Physics test

Read the notes for this chapter

Ways to practise this chapter

Physics · Heat and thermodynamics: questions students ask

How many Physics · Heat and thermodynamics MCQs are there?

14, each with the answer and a short explanation, across 1 topics: Kinetic theory and thermodynamics. Every chapter test draws ten of them, so you can take it many times.

Is the NUST NET Physics chapter 4 test free?

Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.

How is the chapter test marked?

The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.

Can I read the explanations in Urdu?

Yes. In practice mode and on your results you can switch the explanations to Urdu.

Keep going

← Chapter 3: Physics · Oscillations, waves and lightChapter 5: Physics · Electricity →All chaptersStuck? Ask a question